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Find the values for k so that the intersection of x = 2k and 3x + 2y = 12 lies in the first quadrant.

Respuesta :

mhanifa

Answer:

0 < k < 2

Step-by-step explanation:

In the first quadrant both of x and y get positive values, so

  • x > 0 and y > 0
  • x = 2k, k > 0

And replacing x with 2k in the second equation:

  • 3x + 2y = 12
  • 3*2k + 2y = 12
  • 2y= 12 - 6k
  • y = 6 - 3k

Since y > 0:

  • 6 - 3k> 0
  • 2 - k > 0
  • k < 2

Combining both k > 0 and k < 2, we get:

  • 0 < k < 2
sqdancefan

Answer:

  0 < k < 2

Step-by-step explanation:

The intersection will lie in the first quadrant when the solution has the characteristics: x > 0, y > 0.

For x > 0, we require ...

  x = 2k

  x > 0

  2k > 0 . . . . substitute for x

  k > 0 . . . . . divide by 2

__

For y > 0, we require ...

  y = (12 -3x)/2

  y = (12 -3(2k))/2 = 6 -3k . . . . . substituted for x and simplify

  y > 0

  6 -3k > 0 . . . . . . substitute for y

  2 -k > 0 . . . . . . . divide by 3

  k < 2 . . . . . . . . . add k

So, the values of k that result in the intersection being in the first quadrant are ...

  0 < k < 2

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